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OverLord2011 [107]
3 years ago
15

The recommended daily sugar intake is 50.0 gram. How many moles of sugar (C12H22O11) are in 50.0 grams of sugar?

Chemistry
1 answer:
Kitty [74]3 years ago
3 0

Answer:

0.14607220033346532

Explanation:

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Show that 1 kJ/kg = 1000 m2/S2
Sophie [7]

Answer:

1000m2 / s2

Explanation:

Hello! In order to verify this, we have to do unit conversion. We also have to know that J (Joule) = kg * m2 / s2

Then we can start with the test.

1kJ / kg * (1000J / 1kJ) = 1000J / kg

1000J / kg = 1000kg * m2 / kg * s2

In this step we can simplify "kg".

So the result is

1000m2 / s2

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3 years ago
(I WILL MARK BRAINLIEST) What part of the wave does “d” represent?
love history [14]
The answers tot the question above is amplitude 
8 0
3 years ago
The electrons that are in the outermost electron shell, also known as the ____________ , have ____________ energy than those in
Tamiku [17]
The electrons that are in the outermost electron shell, also known as the valence shell, have more energy then those in the inner electron shells, I hope this helps<3
8 0
2 years ago
Can someone please please help
Llana [10]

Answer:

oxidizer

Explanation:

an example of an oxidizers are oxygen and hydrogen peroxide

7 0
3 years ago
Answer the following questions about the fermentation of glucose (C6H12O6, molar mass 180.2 g/mol)
Yuri [45]

The amount of energy in kilocalories released from 49 g of glucose given the data is -4.4 Kcal

How to determine the mole of glucose

Mass of glucose = 49 g

Molar mass of glucose = 180.2 g/mol

Mole of glucose = ?

Mole = mass / molar mass

Mole of glucose = 49 / 180.2

Mole of glucose = 0.272 mole

How to determine the energy released

C₆H₁₂O₆ →2C₂H₆O + 2CO₂  ΔH = -16 kcal/mol

From the balanced equation above,

1 mole of glucose released -16 kcal of energy

Therefore,

0.272 mole of glucose will release = 0.272 × -16 = -4.4 Kcal

Thus, -4.4 Kcal were released from the reaction

Learn more about stoichiometry:

brainly.com/question/14735801

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1 year ago
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